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Question

# The value of $\left(1+\mathrm{i}{\right)}^{5}\left(1-\mathrm{i}{\right)}^{5}$

A

$-8$

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B

$8\mathrm{i}$

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C

$8$

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D

$32$

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Solution

## The correct option is D $32$Explanation for the correct option.Find the valueGiven: $\left(1+\mathrm{i}{\right)}^{5}\left(1-\mathrm{i}{\right)}^{5}$ $={\left(1+\mathrm{i}\right)}^{4}\left(1+\mathrm{i}\right){\left(1-\mathrm{i}\right)}^{4}\left(1-\mathrm{i}\right)-\left(1\right)$${\left(1+\mathrm{i}\right)}^{2}={1}^{2}+2×1×\mathrm{i}+{\mathrm{i}}^{2}=1+2\mathrm{i}-1=2\mathrm{i}\left[{\left(\mathrm{a}+\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\right]\phantom{\rule{0ex}{0ex}}{\left(1-\mathrm{i}\right)}^{2}={1}^{2}-2×1×\mathrm{i}+{\mathrm{i}}^{2}=1-2\mathrm{i}-1=-2\mathrm{i}\left[{\left(\mathrm{a}+\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\right]$Put the above value in (1)$⇒{\left(2\mathrm{i}\right)}^{2}\left(1+\mathrm{i}\right){\left(-2\mathrm{i}\right)}^{2}\left(1-\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}=4{\mathrm{i}}^{2}×4{\mathrm{i}}^{2}×\left(1-{\mathrm{i}}^{2}\right)\left[\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)=\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}=-4×-4×\left(1+1\right)\left[{\mathrm{i}}^{2}=-1\right]\phantom{\rule{0ex}{0ex}}=16×2\phantom{\rule{0ex}{0ex}}=32$Hence $o\mathrm{ption}\left(\mathrm{D}\right)$ is the correct answer.

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